Question

Evaluate the definite integral: \[ I = \int_{0}^{\frac{\pi}{2}} (\sqrt{\tan x} + \sqrt{\cot x})dx \]

• A. \( \frac{\pi}{\sqrt{2}} \)
• B. \( \pi \sqrt{2} \)
• C. \( \frac{\pi}{2} \)
• D. \( \sqrt{2} \pi \)

Hint:

Use integral properties to transform and simplify definite integrals.

Answer 1

The Correct Option is B

Step 1: Expressing the integral. We need to evaluate the integral: \[ I = \int_{0}^{\frac{\pi}{2}} (\sqrt{\tan x} + \sqrt{\cot x}) \, dx \] We will split the integral into two parts: \[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx + \int_{0}^{\frac{\pi}{2}} \sqrt{\cot x} \, dx \] 
Step 2: Evaluating the first integral. Consider the integral \( \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx \). By the substitution \( u = \frac{\pi}{2} - x \), we have \( du = -dx \) and \( \tan \left( \frac{\pi}{2} - u \right) = \cot u \). This gives: \[ \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx = \int_{0}^{\frac{\pi}{2}} \sqrt{\cot u} \, du \] Thus, both integrals \( \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx \) and \( \int_{0}^{\frac{\pi}{2}} \sqrt{\cot x} \, dx \) are equal. 
Step 3: Simplifying the result. Therefore, we can write: \[ I = 2 \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx \] Using known integral results, we know that: \[ \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \, dx = \frac{\pi}{2} \sqrt{2} \] Thus: \[ I = 2 \times \frac{\pi}{2} \sqrt{2} = \pi \sqrt{2} \]